Question

Show the integral of {2arctan(1+a*ln(x))} between 0 and 1 = (pi/2) - a- a^2 - a^3 + O(a^4), where a<<1

Answer 1

\[\int\limits_{0}^{1}2\arctan(1+aln(x)) dx = \Pi/2 - a - a^2 - a^3 + O(a^4)\]

Answer 2

let ln(x) = y => dy = 1/x dx = e^(-y) dx

\[ \int_{-\infty }^0 2\arctan( 1 + a y) e^{y}dy \]
\Change ( y\to-y \) which gines
\[ 2 \int_0^\infty e^{-y} \arctan( 1 - ay) dy = 2 \left[- \arctan(1 -ay)e^{-y} \right ]_0^\infty - 2 \int_0^\infty \frac{a}{1 + (1-ay)^2} e^{-y} dy \]
This gives
\[ \frac \pi 2 - \int_0^\infty \frac{a}{1 + (1-ay)^2} e^{-y} dy \]

Answer 3

\[
2 \int_0^\infty \frac{a}{1 + (1-ay)^2} e^{-y} dy \\
= 2 \int_0^\infty \left(
\frac{\text{ a}}{2}+\frac{\text{ a}^2 \text{ y}}{2}+\frac{\text{ a}^3 \text{ y}^2}{4}-\frac{\text{ a}^5 \text{ y}^4}{8}-\frac{\text{ a}^6 \text{ y}^5}{8}+O\left(\text{ y}^6\right) \right ) e^{-y } dy \\
= a + a^2 + a^3 + \mathscr{O}(a^4)\]

Answer 4

Thank you.

Answer 5

yw